June 2022 Paper 1 Q12
12
(a) A geometric sequence has first term 1 and common ratio \(\dfrac{1}{2}\)
(i) Find the sum to infinity of the sequence. [2 marks]
(ii) Hence, or otherwise, evaluate\[\sum_{n=1}^{\infty} (\sin 30^\circ)^n\] [2 marks]
(b) Find the smallest positive exact value of \(\theta\), in radians, which satisfies the equation\[\sum_{n=0}^{\infty} (\cos\theta)^n = 2 - \sqrt{2}\] [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| (i) Uses \(\dfrac{a}{1 - r}\) | M1 | 1.1a |
| Obtains 2 | A1 | 1.1b |
| (2) | ||
| (ii) Deduces \(a = \dfrac{1}{2}\) and \(r = \dfrac{1}{2}\) or Deduces \(\displaystyle\sum_{n=1}^{\infty}(\sin 30^\circ)^n\) = their part (a)(i) -1 or Deduces the answer is half of their answer in part (a)(i) | M1 | 2.2a |
| Obtains 1 | A1 | 1.1b |
| (2) |
Typical solution
(i)
\[S_\infty = \frac{1}{1 - \frac{1}{2}} = 2\](ii)
\[\sum_{n=1}^{\infty}(\sin 30^\circ)^n = \frac{1}{2} + \frac{1}{4} + \ldots\]\[= \frac{\frac{1}{2}}{1 - \frac{1}{2}}\]\[= 1\]| Scheme | Marks | AO |
|---|---|---|
| Forms equation \(\dfrac{a}{1 - r} = 2 - \sqrt{2}\) If the above is not seen then condone \(\dfrac{\cos\theta}{1 - \cos\theta} = 2 - \sqrt{2}\) or Condone use of a numerical value for \(a\) where \(a \gt 0\) | M1 | 3.1a |
| Uses \(a = 1\) and \(r = \cos\theta\) | B1 | 1.1b |
| Obtains either \(r = 1 - \dfrac{1}{2 - \sqrt{2}}\) or \(-\dfrac{\sqrt{2}}{2}\) or \(\cos\theta = 1 - \dfrac{1}{2 - \sqrt{2}}\) or \(-\dfrac{\sqrt{2}}{2}\) ACF | A1 | 1.1b |
| Deduces \(\theta = \dfrac{3\pi}{4}\) | R1 | 2.2a |
| (4) | ||
| (8 marks) |